Double Integrals · Superchallenge problem

Numbers $x$ and $y$ are chosen independently and uniformly at random from $(0, 1)$. Find the probability that $\left\lfloor \dfrac{1}{x} \right\rfloor = \left\lfloor \dfrac{1}{y} \right\rfloor$. You may use $\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2} = \dfrac{\pi^2}{6}$. Give an exact answer.